A wave pulse travels down a slinky. The mass of the slinky is m = 0.94 kg and is
ID: 1478703 • Letter: A
Question
A wave pulse travels down a slinky. The mass of the slinky is m = 0.94 kg and is initially stretched to a length L = 7.3 m. The wave pulse has an amplitude of A = 0.23 m and takes t = 0.41 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.44 Hz.
1)
What is the speed of the wave pulse?
m/s
2)
What is the tension in the slinky?
N
3)
What is the average speed of a piece of the slinky as a complete wave pulse passes?
m/s
4)
What is the wavelength of the wave pulse?
m
5)
Now the slinky is stretched to twice its length (but the total mass does not change).
What is the new tension in the slinky? (assume the slinky acts as a spring that obeys Hooke’s Law)
N
6)
What is the new mass density of the slinky?
kg/m
7)
What is the new time it takes for a wave pulse to travel down the slinky?
s
8)
If the new wave pulse has the same frequency, what is the new wavelength?
Explanation / Answer
L = 7.3 m
m = 0.94 kg
f = 0.44 Hz
1.) v = distance/time = 7.3 / .41 = 17.80 m/s
2.)T = m*v^2 / L
= 0.94 kg* 17.80 ^2/7.3 = 40.799
3.) 4*amplitude*frequency = 4*0.44*0.23 = 0.4048
4.) wavelength = velocity*period (while period is 1/.44Hz)
= 17.80*1/.44 = 40.45
5.) PAY ATTENTION double the distance new velocity = 2L / .41 = 2*7.3/0.41 = 35.61
6.) m / 2L
7.) t = distance over velocity = 2L / new velocity (31.06m/s)
8.) wave length = period * new velocity = (1/.45Hz) * (31.06m/s)
9.) both frequency and amplitude